2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.
Solution: 
Moles of hydrogen gas = 
Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol
Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l
6M = 
where V is the volume needed

V = 2.083 l
<h3>
What is Titration?</h3>
- Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
- Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
- To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
- Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.
To learn more about titration with the given link
brainly.com/question/2728613
#SPJ4
Answer:
Both have the same amount of particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.
This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.
Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.
Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.